A few families took a trip to an amusement park together. Tickets cost $$7.00$ each for adults and $$2.50$ each for kids, and the group paid $$41.00$ in total. There were $5$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Explanation: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${7x+2.5y = 41}$ ${x = y-5}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-5}$ for $x$ in the first equation. ${7}{(y-5)}{+ 2.5y = 41}$ Simplify and solve for $y$ $ 7y-35 + 2.5y = 41 $ $ 9.5y-35 = 41 $ $ 9.5y = 76 $ $ y = \dfrac{76}{9.5} $ ${y = 8}$ Now that you know ${y = 8}$ , plug it back into ${x = y-5}$ to find $x$ ${x = }{(8)}{ - 5}$ ${x = 3}$ You can also plug ${y = 8}$ into ${7x+2.5y = 41}$ and get the same answer for $x$ ${7x + 2.5}{(8)}{= 41}$ ${x = 3}$ There were $3$ adults and $8$ kids.